What is the derivative of f(x)=(x^2-3x)ln(x^2-3x) ?

The derivative of f(x)=(x^2-3x)ln(x^2-3x) is ln(x^2-3x)(2x-3)+2x-3

What is the first derivative of f(x)=(x^2-3x)ln(x^2-3x) ?

The first derivative of f(x)=(x^2-3x)ln(x^2-3x) is ln(x^2-3x)(2x-3)+2x-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

f (x) = (x^{2} - 3 x) ln (x^{2} - 3 x)

ln (x^{2} - 3 x) (2 x - 3) + 2 x - 3

Solution steps

\frac{d}{d x} ((x^{2} - 3 x) ln (x^{2} - 3 x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2} - 3 x, g = ln (x^{2} - 3 x)

= \frac{d}{d x} (x^{2} - 3 x) ln (x^{2} - 3 x) + \frac{d}{d x} (ln (x^{2} - 3 x)) (x^{2} - 3 x)

\frac{d}{d x} (x^{2} - 3 x) = 2 x - 3

\frac{d}{d x} (ln (x^{2} - 3 x)) = \frac{2 x - 3}{x ^{2} - 3 x}

= (2 x - 3) ln (x^{2} - 3 x) + \frac{2 x - 3}{x ^{2} - 3 x} (x^{2} - 3 x)

\frac{2 x - 3}{x ^{2} - 3 x} (x^{2} - 3 x) = 2 x - 3

= ln (x^{2} - 3 x) (2 x - 3) + 2 x - 3

z = - 3 - 3 i

y = x^{2} + 1

f (x) = - 2 x + 2

ln (3) x^{2}

x = - 4

What is the derivative of f(x)=(x^2-3x)ln(x^2-3x) ?
The derivative of f(x)=(x^2-3x)ln(x^2-3x) is ln(x^2-3x)(2x-3)+2x-3
What is the first derivative of f(x)=(x^2-3x)ln(x^2-3x) ?
The first derivative of f(x)=(x^2-3x)ln(x^2-3x) is ln(x^2-3x)(2x-3)+2x-3